The fourth and seventh terms of an Arithmetic Progression (A.P.) are 60 and 114 respectively. Find the :
(a) first term and common difference.
(b) sum of its first 10 terms.
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(a) Given,
The fourth and seventh terms of an Arithmetic Progression (A.P.) are 60 and 114 respectively.
We know that,
an = a + (n - 1)d
Fourth term :
⇒ a4 = a + (4 - 1)d
⇒ a + 3d = 60 ....(1)
Seventh term :
⇒ a7 = a + (7 - 1)d
⇒ a + 6d = 114 ....(2)
Subtracting equation (1) from equation (2), we get :
⇒ (a + 6d) - (a + 3d) = 114 - 60
⇒ a + 6d - a - 3d = 54
⇒ 3d = 54
⇒ d = $\frac{54}{3}$ = 18.
Substituting value of d = 18 in equation (1), we get :
⇒ a + 3d = 60
⇒ a + 3(18) = 60
⇒ a + 54 = 60
⇒ a = 60 - 54
⇒ a = 6.
Hence, first term = 6 and common difference = 18.
(b) Sum of the first 10 terms :
By formula,
$S_n = \frac{n}{2}[2a + (n - 1)d]$
Substituting values, we get :
$$\begin{aligned}\Rightarrow S_{10} = \frac{10}{2}[2(6) + (10 - 1)18] \\[1em] = 5[12 + 162] \\[1em] = 5[174] \\[1em] = 870.\end{aligned}$$
Hence, sum of its first 10 terms = 870.